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7 months ago

Class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Thirteen

Consider a rectangular block of wood moving with a velocity v₀ in a gas at temperature T and mass density ρ. Assume the velocity is along x-axis and the area of cross-section of the block perpendicular to v₀ is A. Show that the drag force on the block is 4ρ 0 kT Av m , where m is the mass of the gas molecule

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Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

Consider the diagram

let n =number of molecules per unit volume

V_rms= rms speed of gas molecule

When block is moving with speed v_o,relative speed of molecules w.r.t front face =v+v_o

Coming head on, momentum transferred to block per collision =2m (v+v_o)

Number of collisions in time $? t$ = $\frac{1}{2}$ (v+v_o)n $? t$ A where A is the area of cross section.

So momentum transferred in time $? t$ =m (v+v_o)²nA $? t$ this is from front surface

Similarly momentum transferred in time $? t$ = m (v-v_o)²nA $? t$ ) this is from back surface

Drag force = mnA (v+v_o)²- (v-v_o)²)

= mnA (4w_o)=4mnAvv_o

= 4 $ρ A$ vv_o

So $ρ$ =mn/V=M/

...more

This is a long answer type question as classified in NCERT Exemplar

Consider the diagram

let n =number of molecules per unit volume

V_rms= rms speed of gas molecule

When block is moving with speed v_o,relative speed of molecules w.r.t front face =v+v_o

Coming head on, momentum transferred to block per collision =2m (v+v_o)

Number of collisions in time $? t$ = $\frac{1}{2}$ (v+v_o)n $? t$ A where A is the area of cross section.

So momentum transferred in time $? t$ =m (v+v_o)²nA $? t$ this is from front surface

Similarly momentum transferred in time $? t$ = m (v-v_o)²nA $? t$ ) this is from back surface

Drag force = mnA (v+v_o)²- (v-v_o)²)

= mnA (4w_o)=4mnAvv_o

= 4 $ρ A$ vv_o

So $ρ$ =mn/V=M/V

If v = velocity along x axis

Then we can write KE= ½ mv²=1/2 K_BT

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New answer posted

7 months ago

Class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Thirteen

A box of 1.00m³ is filled with nitrogen at 1.50 atm at 300K. The box has a hole of an area 0.010 mm². How much time is required for the pressure to reduce by 0.10 atm, if the pressure outside is 1 atm.

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Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

Given volume V = 1m³

area = 0.01mm²

= 8.01 $* 10^{- 6}$ m²= $10^{- 8}$ m²

Temperature both inside and outside are equal

So initial pressure inside the box = 1.50atm

Final pressure inside the box= 0.1atm

Assuming V_x= speed of nitrogen molecule in x direction

n_i = number of molecules per unit volume in a time interval of $? T$

Let area of the wall, number of particles colliding in time

$? t$ = $\frac{1}{2} n$ _i (v_{x
$? t$})A , here we use ½ because particle moves both in positive and negative direction.

V_x²+ V_y²+ V_z²= V_rms²

V_x²= V_rms²/3 if all three velocities are equal.

½ mv_rms²= 3/2K_BT/m

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7 months ago

Class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Thirteen

Ten small planes are flying at a speed of 150 km/h in total darkness in an air space that is 20 * 20 * 1.5 km³ in volume. You are in one of the planes, flying at random within this space with no way of knowing where the other planes are. On the average about how long a time will elapse between near collision with your plane. Assume for this rough computation that a safety region around the plane can be approximated by a sphere of radius 10m.

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Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

Time require to avoid the collision T= l/v where l = mean free path =1/ $\sqrt[]{2} π d^{2} n$

Where n = N/V

n=number of aeroplanes/volume

= $\frac{10}{20 * 20 * 1.5} = 0.0167 k m$ ^-3

T= $\frac{1 V}{\sqrt[]{2} π d^{2} N} * \frac{1}{v}$

T= $\frac{1}{\sqrt[]{2} * 3.14 * 20^{2} * 0.0167 * 10^{- 6} * 150}$ = $\frac{10^{6}}{1776.25 * 2.505}$

$= \frac{10^{6}}{4449.5} = 224.74 h \approx 225 h$

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New answer posted

7 months ago

Class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Thirteen

Consider an ideal gas with the following distribution of speeds

(a) Calculate V_rms and hence T(m=3 $* 10^{- 26} k g$ )

(b) If all the molecules with speed 1000m/s escape from the system , calculate new V_rms and hence T.

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Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

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New answer posted

7 months ago

Class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Thirteen

Explain why

(a) There is no atmosphere on moon.

(b) There is fall in temperature with altitude. 13.28 Consider an ideal gas with following distribution of speeds

(i) Calculate Vrms and hence T. 26 ( 3.0 10 kg) m − = *

(ii) If all the molecules with speed 1000 m/s escape from the system, calculate new V_rms and hence T.

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Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

(a) The moon has small gravitational force and hence, the escape velocity is small .

As the moon in the proximity of the earth as seen from the sun, the moonhas the same amount of heat per unit area as that of the earth.

The air molecules have large range of speeds . even though the rms speed of the air molcules is smaller than the escape velocity on the moon, a significant number of molecules have speed grater than the escape velocity.

(b) As the molecules move higher their potential energy increases and hence kinetic energy decreases and hence temperature reduces. At g

...more

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7 months ago

Class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Twelve

Consider a heat engine as shown in Fig. Q₁and Q₂ are heat added to heat bath T₁ and heat taken from T₂ in one cycle of engine. W is the mechanical work done on the engine. If W > 0, then possibilities are:

(a) Q₁> Q₂> 0

(b) Q₂ > Q₁ > 0

(c) Q₂ < Q₁ < 0

(d) Q₁ < 0, Q₂ > 0

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a), (c) Q₁= W+Q₂

W=Q₁-Q₂>0

Q₁>Q₂>0

We can also write Q₂1

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New answer posted

7 months ago

Class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Twelve

Consider a cycle followed by an engine 1 to 2 is isothermal 2 to 3 is adiabatic 3 to 1 is adiabatic Such a process does not exist because

(a) Heat is completely converted to mechanical energy in such a process, which is not possible.

(b) Mechanical energy is completely converted to heat in this process, which is not possible.

(c) Curves representing two adiabatic processes don't intersect.

(d) Curves representing an adiabatic process and an isothermal process don't intersect.

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a) the given process is a cyclic process i.e returns to the original state 1

Hence change in internal energy dU =0

dQ= dU+dW=0+dW

hence total heat supplied is converted to work done by the gas which is not possible by second law of thermodynamics.

(c) When the gas expands adiabatically from 2 to 3 . it is not possible to return to the same state without being heat supplied hence 3 to 1 cannot be adiabatic.

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New answer posted

7 months ago

Class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Twelve

Figure 12.5 shows the P-V diagram of an ideal gas undergoing a change of state from A to B. Four different parts I, II, III and IV as shown in the figure may lead to the same change of state.

(a) Change in internal energy is same in IV and III cases, but not in I and II

(b) Change in internal energy is same in all the four cases

(c) Work done is maximum in case I

(d) Work done is minimum in case II

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(b), (c) Change in internal energy for process A to B

dU=nC_vdT=nC_v (dT)=nC_v (T_B-T_A)

work done from A to B = area under the PV curve which is maximum for path I

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New answer posted

7 months ago

Class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Twelve

An ideal gas undergoes isothermal process from some initial state i to final state f. Choose the correct alternatives.

(a) dU = 0

(b) dQ= 0

(c) dQ = dU

(d) dQ = dW

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a), (d) For isothermal dT= 0 so T=constant

For an ideal gas dU = change in internal energy = nC_vdT=0

From first law of thermodynamics dQ= dU+dW

dQ= dW

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New answer posted

7 months ago

Class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Twelve

Which of the processes described below are irreversible?

(a) The increase in temperature of an iron rod by hammering it

(b) A gas in a small container at a temperature T₁ is brought in contact with a big reservoir at a higher temperature T₂ which increases the temperature of the gas

(c) A quasi-static isothermal expansion of an ideal gas in cylinder fitted with a frictionless piston

(d) An ideal gas is enclosed in a piston cylinder arrangement with adiabatic walls. A weight W is added to the piston, resulting in compression of gas

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a), (b), (d) When the rod is hammered the external work is done on the rod which increases its temperature.

Heat is transferred to the gas in the small container by big reservoir at temperature T₂

As the weight is added to the cylinder arrangement in the form of external pressure so it cannot reversed.

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